Respuesta :
Answer:
-5/3 ft/sec
Step-by-step explanation:
In this question, we are asked to calculate the rate at which the height of the top of a ladder is changing given that the lower end is being dragged at a particular rate.
Please check attachment for complete solution and step by step explanation.
Answer:
The rate of change of height, dy / dt = -1.667 ft/s
Step-by-step explanation:
Solution:-
- The length of the ladder, L = 39 ft
- The foot of the ladder is moved away from wall at a rate, dx/dt = 4ft/s
Find:-
how fast is the height of the top changing (this will be a negative rate) when the lower end is 15 feet from the wall?
Solution:-
- We will first draw a right angle triangle, with vertical height of the ladder to be"y" and distance of the foot of the ladder and the wall to be "x".
- Then express the length "L" in terms of x and y using pythagorean theorem:
L^2 = x^2 + y^2
y^2 = 39^2 - x^2
- Taking height of the ladder as the dependent variable and distance of the foot of the ladder from wall as independent variable.
- Formulate a differential equation from the given expression above in terms of "dy/dt" and "dx/dt". Perform implicit differential of the computed expression "d/dt":
2y*dy/dt = -2x*dx/dt
dy / dt = -(x/y)*dx/dt
- Where, dy / dt : The change in height of the ladder.
- The height of the ladder at x = 15 ft is:
[tex]y = \sqrt{39^2 - x^2} = \sqrt{39^2 - 15^2} = 36[/tex]
- Then evaluate dy/dt:
dy / dt = -(15/36)*4
dy / dt = -1.667 ft/s
